College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.3 - Complex Numbers; Quadratic Equations in the Complex Number System - 1.3 Assess Your Understanding: 48

Answer

$0$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ i^7+i^5+i^3+i ,$ use the laws of exponents and the equivalence $i^2=-1.$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} i^{6+1}+i^{4+1}+i^{2+1}+i \\\\= i^6\cdot i^1+i^4\cdot i^1+i^{2}\cdot i^1+i \\\\= i^6\cdot i+i^4\cdot i+i^{2}\cdot i+i .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} i^{2\cdot3}\cdot i+i^{2\cdot2}\cdot i+i^{2}\cdot i+i \\\\ (i^2)^3\cdot i+(i^2)^2\cdot i+i^{2}\cdot i+i .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (-1)^3\cdot i+(-1)^2\cdot i+(-1)\cdot i+i \\\\= (-1)\cdot i+(1)\cdot i+(-1)\cdot i+i \\\\= -i+i-i+i \\\\= 0 .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.